If $x \bigtriangleup y = x^{2}-4y^{2}$ and $x \star y = 3x+y$, find $(-2 \star 4) \bigtriangleup 0$.
First, find $-2 \star 4$ $ -2 \star 4 = (3)(-2)+4$ $ \hphantom{-2 \star 4} = -2$ Now, find $-2 \bigtriangleup 0$ $ -2 \bigtriangleup 0 = (-2)^{2}-4(0^{2})$ $ \hphantom{-2 \bigtriangleup 0} = 4$.